\[\begin{align}
& \text{if }f\left( \alpha \right)=x\cos \alpha +y\sin \alpha -p\left( \alpha \right),\text{ } \\
& \text{then the lines }f\left( \alpha \right)=0\text{ and }f\left( \beta \right)=0\text{ are }\bot \text{ to each other, if} \\
& \left( a \right)\text{ }\alpha =\beta \text{ }\left( b \right)\alpha +\beta =\dfrac{\pi }{2} \\
& \left( c \right)|\alpha -\beta |=\dfrac{\pi }{2}\text{ }\left( d \right)\alpha \pm \beta =\dfrac{\pi }{2} \\
\end{align}\]
Answer
646.2k+ views
Hint: first find the two equations, by substituting the variables into given function, by substituting first variable you get first line equation similarly when you substitute the second variable you get the second line equation and then apply the condition of perpendicular lines. For finding slope of the line use this formula. If the slope of ax + by + c = 0 is p. Then the value of p is given by:
\[p=-\dfrac{a}{b}\]
If two straight lines with slope a, b are perpendicular then:
\[a\times b=-1\].
Use the given trigonometric identity for simplification purposes:
\[\cos \left( A-B \right)=\cos A.\cos B+\sin A.\sin B\]
Complete step-by-step answer:
So for the first equation the function is already given.
By equating it to 0, we get:
\[x\cos \alpha +y\sin \alpha -p\left( \alpha \right)=0.....\left( 1 \right)\]
The equation (1) becomes the first straight line.
Let the slope of straight line 1 is m.
Slope condition:
If the slope of ax + by + c = 0 is p. Then the value of p is given by:
\[p=-\dfrac{a}{b}.....\left( 2 \right)\]
So by seeing at equation (2) the values of equation (1) are:
p = m
\[a=\cos \alpha \]
\[b=\sin \alpha \]
By substituting the values of p, a, b in equation (2), we get:
\[m=-\dfrac{\cos \alpha }{\sin \alpha }=-\cot \alpha .....\left( 3 \right)\]
So for the second equation the function is already given.
By equating it to 0 and substituting variable, we get:
\[x\cos \beta +y\sin \beta -p\left( \beta \right)=0.....\left( 4 \right)\]
The equation (4) becomes the second straight line.
Let the slope of straight line 2 is n.
Slope condition:
If the slope of ax + by + c = 0 is p. Then the value of p is given by:
\[p=-\dfrac{a}{b}.....\left( 2 \right)\]
So by seeing at equation (2) the values of equation (4) are:
p = n
\[a=\cos \beta \]
\[b=\sin \beta \]
By substituting the values of p, a, b in equation (2), we get:
\[n=-\dfrac{\cos \beta }{\sin \beta }=-\cot \beta .....\left( 5 \right)\]
The condition of perpendicular lines:
If two straight lines with slope a, b are perpendicular then
\[a\times b=-1\].
By applying the above condition, we get:
m.n = -1
\[\left( -\dfrac{\cos \alpha }{\sin \alpha } \right).\left( -\dfrac{\cos \beta }{\sin \beta } \right)=-1\]
By cross multiplying, we get:
\[\cos \alpha .\cos \beta =-\sin \alpha .\sin \beta \]
\[\cos \alpha .\cos \beta +\sin \alpha .\sin \beta =0\]
We know that:
\[\cos \left( A-B \right)=\cos A.\cos B+\sin A.\sin B\]
By applying above trigonometric identity, we get:
\[\cos \left( \alpha -\beta \right)=0\]
\[|\alpha -\beta |={{\cos }^{-1}}\left( 0 \right)=\dfrac{\pi }{2}\]
\[|\alpha -\beta |=\dfrac{\pi }{2}\]
So, the correct answer is “Option c”.
Note: Take care of negative signs properly. As there is already a negative sign in the slope already if you confuse it may lead to the wrong answer. While substituting the second variable you must be careful that you must substitute it in x, y both terms. In a hurry, generally students forget to substitute in one of the terms and then lead to the wrong answer. It does not matter if you don’t substitute in the constant because we will calculate slope and then check perpendicular condition which doesn’t depend on the constant term.
\[p=-\dfrac{a}{b}\]
If two straight lines with slope a, b are perpendicular then:
\[a\times b=-1\].
Use the given trigonometric identity for simplification purposes:
\[\cos \left( A-B \right)=\cos A.\cos B+\sin A.\sin B\]
Complete step-by-step answer:
So for the first equation the function is already given.
By equating it to 0, we get:
\[x\cos \alpha +y\sin \alpha -p\left( \alpha \right)=0.....\left( 1 \right)\]
The equation (1) becomes the first straight line.
Let the slope of straight line 1 is m.
Slope condition:
If the slope of ax + by + c = 0 is p. Then the value of p is given by:
\[p=-\dfrac{a}{b}.....\left( 2 \right)\]
So by seeing at equation (2) the values of equation (1) are:
p = m
\[a=\cos \alpha \]
\[b=\sin \alpha \]
By substituting the values of p, a, b in equation (2), we get:
\[m=-\dfrac{\cos \alpha }{\sin \alpha }=-\cot \alpha .....\left( 3 \right)\]
So for the second equation the function is already given.
By equating it to 0 and substituting variable, we get:
\[x\cos \beta +y\sin \beta -p\left( \beta \right)=0.....\left( 4 \right)\]
The equation (4) becomes the second straight line.
Let the slope of straight line 2 is n.
Slope condition:
If the slope of ax + by + c = 0 is p. Then the value of p is given by:
\[p=-\dfrac{a}{b}.....\left( 2 \right)\]
So by seeing at equation (2) the values of equation (4) are:
p = n
\[a=\cos \beta \]
\[b=\sin \beta \]
By substituting the values of p, a, b in equation (2), we get:
\[n=-\dfrac{\cos \beta }{\sin \beta }=-\cot \beta .....\left( 5 \right)\]
The condition of perpendicular lines:
If two straight lines with slope a, b are perpendicular then
\[a\times b=-1\].
By applying the above condition, we get:
m.n = -1
\[\left( -\dfrac{\cos \alpha }{\sin \alpha } \right).\left( -\dfrac{\cos \beta }{\sin \beta } \right)=-1\]
By cross multiplying, we get:
\[\cos \alpha .\cos \beta =-\sin \alpha .\sin \beta \]
\[\cos \alpha .\cos \beta +\sin \alpha .\sin \beta =0\]
We know that:
\[\cos \left( A-B \right)=\cos A.\cos B+\sin A.\sin B\]
By applying above trigonometric identity, we get:
\[\cos \left( \alpha -\beta \right)=0\]
\[|\alpha -\beta |={{\cos }^{-1}}\left( 0 \right)=\dfrac{\pi }{2}\]
\[|\alpha -\beta |=\dfrac{\pi }{2}\]
So, the correct answer is “Option c”.
Note: Take care of negative signs properly. As there is already a negative sign in the slope already if you confuse it may lead to the wrong answer. While substituting the second variable you must be careful that you must substitute it in x, y both terms. In a hurry, generally students forget to substitute in one of the terms and then lead to the wrong answer. It does not matter if you don’t substitute in the constant because we will calculate slope and then check perpendicular condition which doesn’t depend on the constant term.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

Two of the body parts which do not appear in MRI are class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

What will happen if the mucus is not secreted by the class 11 biology CBSE

